False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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FALSE: convergence in measure implies almost-everywhere convergence

Statement refuted

convergence in measure implies almost-everywhere convergence.

Facts & Assumptions

Given: Lebesgue measure on $[0, 1]$ and the dyadic typewriter sequence $f_{n}$ defined by $f_{0} : = 0$ and $f_{2^{k} + j} : = χ_{I_{k, j}} for k \geq 0, 0 \leq j < 2^{k},$ where $I_{k, j} = [j 2^{- k}, (j + 1) 2^{- k})$ for $j < 2^{k} - 1$ and $I_{k, 2^{k} - 1} = [1 - 2^{- k}, 1]$ .

[L1]

Convergence in measure means that for every real $ε > 0$ , $μ ({∣ f_{n} - f ∣ > ε}) \to 0$ . (Convergence in measure)

[L2]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

Refutation

technique · direct

1.1

If $2^{k} \leq n < 2^{k + 1}$ , then $f_{n}$ is the indicator of an interval of length $2^{- k}$ . Hence for every $ε \in (0, 1)$ , $λ ({∣ f_{n} - 0 ∣ > ε}) = 2^{- k} \to 0,$ so $f_{n} \to 0$ in measure by [L1].

givenL1algebra

2.1

Fix $x \in [0, 1]$ . In each dyadic generation there is exactly one interval containing $x$ , so $f_{n} (x) = 1$ infinitely often; the same generation also contains intervals missing $x$ , so $f_{n} (x) = 0$ infinitely often. Therefore $(f_{n} (x))$ has no limit for any $x \in [0, 1]$ , and [L2] fails.

step 1.1L2

3.1

This refutes the claim.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Section 2.4, Example (iv) (standard reference, not scraped)
Terence Tao, 245A Notes 4: Modes of convergence, Example 7 (standard reference, not scraped)